Let \(t> 1\)  and take nonnegative \(\phi _t\in C^\infty (\mathbb {R})\) , which satisfies:
1). \(\phi _t\)  is 1 on \([\frac{1}{t},t]\)  and 0
on \((-\infty ,\frac{1}{2t}]\cup [2t,+\infty )\) ; 2) \(|\phi ^{\prime }|<2t\) .
Let
\(\hat{g}_{k,t}=\phi _t(|x|) g_k+(1-\phi _t(|x|))g_{euc}.\) 
We have
\(\int _{B_{2t}\setminus B_{t}}|\nabla ({\hat{g}}_{k,t}-g_{euc})|^pdx&\le &Ct^p\int _{B_{2t}\setminus B_{t}}|\hat{g}_{k,t}-g_{euc}|^pdx+C\int _{B_{2t}\setminus B_{t}}|\nabla (\hat{g}_{k,r}-g_{euc})|^pdx\\&\le &C(t)\Vert \hat{g}_{k,t}-g_{euc}\Vert ^p_{W^{1,p}(B_{2t}\setminus B_{t})}.\) 
In the same way, we get a similar estimate on \(B_\frac{1}{t}\setminus B_\frac{1}{2t}\) .
Note that
\(det(\phi _tg_k+(1-\phi _t)g_{euc})\ge \frac{1}{2}\min \lbrace det(g_k),det(g_{euc})\rbrace ,\) 
which implies that
\(\frac{1}{det(\hat{g}_{k,t})}\le C(1+\frac{1}{det(g_k)})=C(1+det(g_k^{-1})).\) 
Since \(g_k^{-1}\)  converges in \(W^{1,p}_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\)  for
any \(p\in (n-1,n)\) , \((g_{k}^{-1})\)  is bounded in \(L^q(B_r\setminus B_\frac{1}{r})\)  for any \(q\) . Then \(1/det(\hat{g}_{k,t})\)  is bounded in \(L^q(B_r\setminus B_\frac{1}{r})\)  for any \(q\) . Recall that
the matrix inverse is just the adjugate matrix divided by the determinant.
We obtain similar
estimates of the \(W^{1,p}\)  norm of \(\hat{g}^{-1}_{k,t}\) .
Since \(\hat{g}_{k,t}=g_{ecu}\)  on \(B_{2t}^c\) , it is not difficult to check that
any geodesic between two points \(x\) , \(y\in B_r\)  must in \(B_{3r}\)  when \(k\) 
is sufficiently large.
Then, using the arguments in [1]}, a subsequence of \(d_{\hat{g}_{k,t}}\)  converges to \(d_{g_{euc}}\) 
in \(C^0(B_r\times B_r)\)  for any \(r\) .
Thus, after passing to a subsequence, we can find \(t_k\rightarrow +\infty \) , such that \(d_{\hat{g}_{k,t_k}}\) 
converges to \(d_{g_{euc}}\)  in \(C^0(B_r\times B_r)\)  for any \(r\) . For simplicity, we set \(\tilde{g}_k=g_{k,t_k}\) .
Next, we prove the \(d_{g_k}\)  converges to \(d_{g_{euc}}\)  in \(C^0((B_r\setminus B_\frac{1}{r})\times (B_r\setminus B_\frac{1}{r}))\) .
Assume this is not true, then there exists \(r\) , and \(x_k\) , \(y_k\in B_r\setminus B_\frac{1}{r}\) ,
such that
\(|d_{g_k}(x_k,y_k)-d_{g_{euc}}(x_k,y_k)|\ge \epsilon >0.\) 
We may assume \(x_k\rightarrow x_\infty \)  and \(y_k\rightarrow y_\infty \) .
We claim that \(d_{g_k}(x_k,x_\infty )\rightarrow 0\) . For a fixed \(\delta \)  with
\(B_\delta (x_\infty )\subset B_r\setminus B_\frac{1}{r}\) , we take \(x_k^{\prime }\in B_\delta (x_\infty )\) , such that \(x_k\)  is on the segment \(\overline{x_\infty x_k^{\prime }}\) .
Let \(\gamma _k(t)=tx_\infty +tx_k^{\prime }\) . Then
\(\limsup _{k\rightarrow +\infty }d_{g_k}(x_k,x_\infty )\le \limsup _{k\rightarrow +\infty }\int _0^1\sqrt{g_{k}(\dot{\gamma }_k,\dot{\gamma }_k)}dt\le \limsup _{k\rightarrow +\infty }(\int _0^1g_k(\dot{\gamma }_k,\dot{\gamma }_k)dt)^2.\) 
By the trace embedding theorem, it is easy to check that \(\int _0^1g_k(\dot{\gamma }_k,\dot{\gamma }_k)dt\rightarrow \delta \) . Letting \(\delta \rightarrow 0\) , we get the claim.
Consequently, we may assume
\(d_{g_k}(x_\infty ,x_k)+d_{g_k}(y_\infty ,y_k)<\frac{\epsilon }{2}.\) 
Then
\(|d_{g_k}(x_\infty ,y_\infty )-d_{g_{euc}}(x_\infty ,y_\infty )|\ge \frac{\epsilon }{2},\) 
when \(k\)  is sufficiently large.
Let \(s\)  be sufficiently large and \(\gamma \subset B_s\setminus B_\frac{1}{s}\)  be a curve from \(x_\infty \) 
to \(y_\infty \) , such that
\(L_{g_{euc}}(\gamma )\le d_{g_{euc}}(x_\infty ,y_\infty )+\epsilon /4\) . By trace
embedding,
\(d_{g_k}(x_\infty ,y_\infty )\le L_{g_k}(\gamma )\rightarrow L_{g_{euc}}(\gamma )\le d_{g_{euc}}(x_\infty ,y_\infty )+\epsilon /4<2r+\frac{\epsilon }{4}.\) 
We
let \(\gamma _k\)  be the curve
from \(x_\infty \)  to \(y_\infty \)  such that
\(d_{g_k}(x_\infty ,y_\infty )+\frac{\epsilon }{4}>L(\gamma _k,g_k).\) 
We claim that \(\gamma _k\subset B_{3r}\)  when \(k\) 
is sufficiently large. Otherwise,
\(d_{g_k}(x_\infty ,y_\infty )+\frac{\epsilon }{4}>L(\gamma _k,g_k)\ge 2d_{g_k}(\partial B_r,\partial B_{3r}).\) 
It is easy to check that
\(d_{g_k}(\partial B_r,\partial B_{3r})=d_{\tilde{g}_k}(\partial B_r,\partial B_{3r})\rightarrow 2r.\) 
Using (REF ), we get
\(4r<2r+\frac{\epsilon }{4}\) 
which is impossible when \(r\)  is selected to be enough large.
The rest of the proof can be divided into 2 cases. Case 1, we assume \(\gamma _k\cap \partial B_\frac{1}{t_k}=\emptyset \) .
In this case,
\(L_{g_k}(\gamma _k)=L_{\tilde{g}_k}(\gamma _k)\ge d_{\tilde{g}_k}(x_\infty ,y_\infty ).\) 
Case 2, we may assume \(x_k^{\prime }\)  and \(y_k^{\prime }\)  to be the first and the last point in \(\gamma _k\cap \partial B_\frac{1}{t_k}\) . Then
\(d_{g_k}(x_\infty ,y_\infty )+\frac{\epsilon }{4}&\ge & L(\gamma _k,g_k)\\&\ge & L(\gamma _k|_{[x_\infty ,x_k^{\prime }]},g_k)+ L(\gamma _k|_{[y^{\prime },y_\infty ]},g_k)\\&=&L(\gamma _k|_{[x_\infty ,x_k^{\prime }]},\tilde{g}_k)+ L(\gamma _k|_{[y^{\prime },y_\infty ]},\tilde{g}_k)\\&\ge &d_{\tilde{g}_k}(x_\infty ,x_k^{\prime })+d_{\tilde{g}_k}(y_k^{\prime },y_\infty )\\&\ge & d_{\tilde{g}_k}(x_\infty ,y_\infty )-d_{\tilde{g}_k}(x_k^{\prime },y_k^{\prime }).\) 
Thus, for both cases, we have
\(\limsup _{k\rightarrow +\infty }d_{g_k}(x_\infty ,y_\infty )+\frac{\epsilon }{4}\ge d_{g_{euc}}(x_\infty ,y_\infty ).\) 
Combining (REF ), we get
\(\limsup _{k\rightarrow +\infty }|d_{g_k}(x_\infty ,y_\infty )-d_{g_{euc}}(x_\infty ,y_\infty )|\le \frac{\epsilon }{4},\) 
which contradicts our assumption.
\(\hfill \Box \)

Three Circles Theorem
In this section, we present Three Circles Theorem. Since it is convenient to state and prove this theorem
on pipes, we let \(Q=[0,3L]\times S^{n-1}\) , and
\(Q_i=[(i-1)L,iL]\times \mathbb {S}^{n-1},\,\,\,\,i=1,2,3.\) 
Set \(g_Q=dt^2+g_{\mathbb {S}^{n-1}}\)  and \(dV_Q=dV_{g_Q}\) .
We first state this theorem for case of \(g=g_Q\)  and \(R=(n-1)(n-2)\) :
Lemma 2.4 
Let \(u\ne 0\)  solve the following equation on \(Q\) :
\(-\Delta u+\frac{(n-2)^{2}}{4}u=0,\) 
then there exists \(L_0\) , such that for any \(L>L_0\) , we have
1) 
\(\displaystyle {\int }_{Q_1}u^2dV_{Q}\le e^{-L}\displaystyle {\int }_{Q_2}u^2dV_{Q}\)  implies
\(\int _{Q_2}u^2dV_{Q}< e^{-L}\int _{Q_3}u^2dV_{Q}.\) 
2) 
\(\displaystyle {\int }_{Q_3}u^2dV_{Q}\le e^{-L}\displaystyle {\int }_{Q_2}u^2dV_{Q}\)  implies
\(\int _{Q_2}u^2dV_{Q}< e^{-L}\int _{Q_1}u^2dV_{Q}.\) 
3) 
either
\(\int _{Q_2}u^2dV_{Q} < e^{-L}\int _{Q_1}u^2dV_{Q}\,\,\,\,or\,\,\,\,\int _{Q_2}u^2dV_{Q}< e^{-L}\int _{Q_3}u^2dV_{Q}.\) 


For the proof, one can refer to [2]}.
Next, we discuss the general case.
Theorem 2.5 
Let \(g_0\)  be a metric over \(Q\)  and \(u\in W^{2,p}\) 
which solves the equation
\(-\Delta _{g_0}u+c(n)R(g_0)u=fu.\) 
There exist \(\epsilon _0^{\prime }\) , \(\tau \) , and \(L_0\) , such that
if
\(\Vert g_Q-g_0\Vert _{C^2(Q)}<\tau ,\,\,\,\,\int _Q|f|^\frac{n}{2}dV_{g_0}<\epsilon _0^{\prime },\) 
then for any \(L>L_0\) , we have
1) 
\(\displaystyle {\int }_{Q_1}u^2dV_{g_0}\le e^{-L}\displaystyle {\int }_{Q_2}u^2dV_{g_0}\)  implies
\(\int _{Q_2}u^2dV_{g_0}\le e^{-L}\int _{Q_3}u^2dV_{g_0}.\) 
2) 
\(\displaystyle {\int }_{Q_3}u^2dV_{g_0}\le e^{-L}\displaystyle {\int }_{Q_2}u^2dV_{g_0}\)  implies
\(\int _{Q_2}u^2dV_{g_0}\le e^{-L}\int _{Q_1}u^2dV_{g_0}.\) 
3) 
either
\(\int _{Q_2}u^2dV_{g_0} \le e^{-L}\int _{Q_1}u^2dV_{g_0}\,\,\,\,or\,\,\,\,\int _{Q_2}u^2dV_{g_0}\le e^{-L}\int _{Q_3}u^2dV_{g_0}.\) 


If the statement in 1) was false, then we can find \(g_k\) ,
\(u_k\)  and \(f_k\) , s.t.
\(g_k\rightarrow g_Q \mbox{ in }C^2,\,\,\,\,\int _{Q}|f_k|^\frac{n}{2}dV_{g_0}\rightarrow 0,\) 
\(-\Delta _{g_k}u_k+c(n)R(g_k)u_k=f_ku_{k},\) 
and
\(\int _{Q_1}|u_k|^2dV_{g_k}\le e^{-L}\int _{Q_2}|u_k|^2dV_{g_k},\,\,\,\,\int _{Q_2}|u_k|^2dV_{g_k} > e^{-L}\int _{Q_3}|u_k|^2dV_{g_k}.\) 
Let
\(v_k=\frac{u_k}{\Vert u_k\Vert _{L^2(Q_2,g_k)}}.\) 
We have
\(\int _{Q_1}|v_k|^2dV_{g_k}\le e^{-L}\int _{Q_2}|v_k|^2dV_{g_k},\,\,\,\,\int _{Q_3}|v_k|^2dV_{g_k} < e^{L}\int _{Q_2}|v_k|^2dV_{g_k},\) 
and
\(\int _{Q_2}|v_k|^2dV_{g_k}=1.\) 
Thus
\(\int _Q|v_k|^2dV_{g_k}\le C.\) 
Since
\(-\Delta _{g_k} v_k+c(n)R(g_k)v_k=f_kv_k,\) 
by Lemma REF  and (REF ), \(v_k\)  converges to a function \(v\)  in \(W^{1,2}_{loc}\) , where \(v\)  satisfies:
\(-\Delta v+\frac{(n-2)^2}{4}v=0,\,\,\,\,and\,\,\,\,\int _{Q_2}|v|^2dV_Q=1.\) 
Thus \(v\ne 0\) .
Moreover,
\(\int _{[\epsilon ,L]\times S^{n-1}}|v|^2dV_Q=\lim _{k\rightarrow +\infty }\int _{[\epsilon ,L]\times S^{n-1}}|v_k|^2dV_Q\le e^{-L}\lim _{k\rightarrow +\infty }\int _{Q_2}|v|^2dV_Q,\) 
letting \(\epsilon \rightarrow 0\)  gives
\(\int _{Q_1}|v|^2dV_Q\le e^{-L}\int _{Q_2}|v|^2dV_Q.\) 
Similarly, there holds
\(\int _{Q_3}|v|^2dV_Q\le e^{L}\int _{Q_2}|v|^2dV_Q,\) 
which contradicts Lemma REF . Hence, the statements in 1) are proved. Using the same arguments, we can easily carry out the proof of 2) and 3).
 \(\hfill \Box \)
Theorem 2.6 
Let \(B\)  be the unit ball of \(\mathbb {R}^n\) , and \(g_0\)  a smooth metric defined on \(\overline{B}\) . \(x^1,\cdots ,x^n\)  are normal coordinates and
\(|g_{0,ij}-\delta _{ij}|\le A|x|^2.\) 
Let \(g=u^\frac{4}{n-4}g_0\)  be a smooth metric
defined on \(\overline{B}\setminus \lbrace 0\rbrace \)  with
\(\int _{B}|R(g)|^\frac{n}{2}dV_g<+\infty ,\,\,\,\,\mathrm {vol}(\overline{B}\setminus \lbrace 0\rbrace ,g)=+\infty .\) 
Then there exists \(r_0\)  and \(L_0\) , such that
for any \(r<r_0\)  and \(L>L_0\) , there holds
\(\int _{B_{r}\setminus B_{rL^{-1}}}\frac{u^2}{|x|^2}dV_{g_0}\le \frac{1}{L}\int _{B_{rL^{-1}}\setminus B_{rL^{-2}}}\frac{u^2}{|x|^2}dV_{g_0}.\) 
Moreover, we have
\(\lim _{k\rightarrow +\infty }\int _{B_{L^{-k}r}\setminus B_{L^{-k-1}r}}\frac{u^2}{|x|^2}dV_{g_0}=\lim _{k\rightarrow +\infty }\int _{B_{L^{-k}r}\setminus B_{L^{-k-1}r}}u^\frac{2n}{n-2}dV_{g_0}=+\infty .\) 

Put
\(\phi (t,\theta )=(e^{-t},\theta ),\) 

and
\(g^{\prime }(t,\theta )=\phi ^*(g)=v^\frac{4}{n-2}\hat{g}(t,\theta ),\) 

where \(\hat{g}(t,\theta )=e^{2t}\phi ^*(g_0)\) , which converges to \(dt^2+g_{\mathbb {S}^{n-1}}\)  as \(t\rightarrow +\infty \) . Then
\(v^\frac{4}{n-2}(t,\theta )=u^\frac{4}{n-2}(e^{-t},\theta )e^{-2t},\) 
\(-\Delta _{\hat{g}}v+c(n)R(\hat{g})v=c(n)R(g^{\prime })v^\frac{n+2}{n-2}:=fv,\) 
\(\int _{S^1\times [a,b]}|f|^\frac{n}{2}dV_{\hat{g}}=c\int _{B_{e^a}\setminus B_{e^b}}|R({g})|^\frac{n}{2}dV_g,\) 

and
\(\int _{B_r\setminus B_{r/L}}\frac{u^2}{r^2}dV_{g_0}=\int _{S^{n-1}\times [-\log r,-\log r+\log L]}v^2dV_{\hat{g}}.\) 
We set
\(\Omega _k=\mathbb {S}^{n-1}\times [-\log r+k\log L,-\log r+(k+1)\log L].\) 
By Theorem REF , if \(\log L\)  is large enough and (REF ) is false, then
\(\int _{\Omega _{k+1}}v^2dV_{\hat{g}}\le CL^{-k},\,\,\,\,\forall k\gg 1,\) 
hence
\(|\lbrace (t,\theta )\in \Omega _k: v(t,\theta )\ge 1\rbrace |\le CL^{-k}.\) 
By Lemma REF  and (REF ), \(\Vert v\Vert _{L^\frac{4n}{n-2}(\Omega _k)}<C\) .
Then
\(\int _{\Omega _k}v^\frac{2n}{n-2}dx&\le &\int _{\Omega _k\cap \lbrace v\le 1\rbrace }v^2+(\int _{\Omega _k\cap \lbrace v\ge 1\rbrace }v^\frac{4n}{n-2})^\frac{1}{2}|\lbrace v\ge 1\rbrace \cap \Omega _k|^\frac{1}{2}\\&\le & C(L^{-k}+L^{-\frac{k}{2}}).\) 
By Lemma REF  and (REF ),
\(\int _{B_r}u^\frac{2n}{n-2}dV_{g_0}\le C\int _{B_r}u^\frac{2n}{n-2}dx= C\sum _k\int _{B_{rL^{-k}}\setminus B_{rL^{-k-1}}}u^\frac{2n}{2n-2}dV_{g_0}\le \sum _k\frac{C}{L^k}<+\infty ,\) 
a contradiction.
Thus, we get (REF ),
which implies that
\(\int _{B_{rL^{-k}}\setminus B_{rL^{-k-1}}}\frac{u^2}{r^2}dx\ge CL^k.\) 
Since
\(\int _{B_{rL^{-k}}\setminus B_{rL^{-k-1}}}\frac{u^2}{r^2}dx\le C(\log L)^\frac{2}{n}(\int _{B_{rL^{-k}}\setminus B_{rL^{-k-1}}}u^\frac{2n}{n-2}dx)^\frac{n-2}{n},\) 
we have
\(\lim _{r\rightarrow 0}\int _{B_{rL^{-k}}\setminus B_{rL^{-k-1}}}u^\frac{2n}{n-2}dV_{g_0}=\infty .\) 
\(\hfill \Box \)
It is easy to check that, if we replace \(\mathrm {vol}(B,g)=+\infty \)  with \(\mathrm {vol}(B,g)<+\infty \) , then
\(\int _{\Omega _k}v^2dx\ge L^{-k}.\) 
Consequently, \(u\in L^p\)  for some \(p>\frac{2n}{n-2}\) , which follows that \(R(g)u^\frac{n+2}{n-2}\in L^q\)  for some \(q>1\) . Then it is not
difficult to get \(u\in W^{2,q}(B)\) , and then \(u^\frac{2}{n-2}\in W^{2,p}(B)\) .
By trace embedding inequality, \((\overline{B}\setminus \lbrace 0\rbrace ,g)\)  is bounded. For more details,
one can refer to [3]}.
asymptotic properties
In the rest of this work, we will use \(B\)  to denote the unit ball of \(\mathbb {R}^n\) ,
and \(g_0\)  a smooth metric defined on \(\overline{B}\) . For simplicity,
we always assume \(x^1,\cdots ,x^n\)  are normal coordinates and
\(|g_{0,ij}-\delta _{ij}|\le c|x|^2.\) 
Moreover, \(g=u^\frac{4}{n-4}g_0\)  always denotes a smooth metric
on \(\overline{B}\setminus \lbrace 0\rbrace \)  with
\(\mathrm {vol}(B\setminus \lbrace 0\rbrace ,g)=\infty .\) 
First of all, we prove the following:
Lemma 3.1 
Let \(r_k\rightarrow 0\) . After passing to a subsequence, we can find \(c_k\) , such that
\(c_kr_k^{\frac{n-2}{2}}u(r_kx)\)  converges to \(|x|^{2-n}\)  weakly in \(W^{2,p}_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\) 
for any \(p\in [1,\frac{n}{2})\) . Moreover, \(d_{g_k}\)  converges to
\(d_{|x|^{-4}g_{euc}}\)  in \(C^0((B_r\setminus B_\frac{1}{r})\times (B_r\setminus B_\frac{1}{r}))\)  for any \(r>1\) , where
\(g_{k,ij}=u_k^\frac{4}{n-2}g_{0,ij}(r_kx)\) .

Put
\(u_k(x)=r_k^{\frac{n-2}{2}}u(r_kx)c_k,\) 
where \(\int _{\partial B_1}\log u_k d{\mathbb {S}^{n-1}}=0\) . It is easy to check that
\(-\Delta _{g_k} u_k=f_ku_k\) , where
\(f_k=-c(n)r_k^2R_{g_0}(r_kx)+c(n)R(r_kx)(r_k^\frac{n-2}{2}u_k(r_kx))^\frac{4}{n-2},\) 
which satisfies, for sufficiently large \(k\) ,
\(\Vert f_k\Vert _{L^\frac{n}{2}(B_r\setminus B_\frac{1}{r})}\le Cr_k^2+C\left(\int _{B_{rr_k}\setminus B_\frac{r_k}{r}}|R|^\frac{n}{2}u^\frac{2}{n}dx\right)^\frac{n-2}{2n}\le \min \lbrace \epsilon _0,\epsilon _0^{\prime }\rbrace .\) 
By Lemma REF  and the Poincaré inequality ( c.f. [4]}),
\(\log u_k\)  is bounded in \(W^{1,2}(B_r\setminus B_\frac{1}{r})\) .
Then, by Lemma REF , \(u_k\)  is bounded in \(W^{2,p}(B_r\setminus B_\frac{1}{r})\) , which follows that \(u_k\)  converges weakly to a positive harmonic function \(u^{\prime }\) 
locally on \(\mathbb {R}^n\setminus \lbrace 0\rbrace \)  with \(\int _{\partial B_1}\log u^{\prime }d\mathbb {S}^{n-1}=0\) .
By Corollary 3.14 in [5]}, \(u^{\prime }\)  is written as
\(u^{\prime }=a+b|x|^{2-n},\) 
where \(a\)  and \(b\)  are nonnegative constants.
Applying Theorem REF , we get
\(\int _{B_{r_kr}\setminus B_{r_krL^{-1}}}\frac{u^2}{|x|^2}dV_{g_0}\le \frac{1}{L}\int _{B_{r_krL^{-1}}\setminus B_{r_krL^{-2}}}\frac{u^2}{|x|^2}dV_{g_0},\) 
which means that
\(\int _{B_r\setminus B_{rL^{-1}}}\frac{u_k^2}{|x|^2}dV_{g_{0,k}}\le \frac{1}{L}\int _{B_{rL^{-1}}\setminus B_{rL^{-2}}}\frac{u_k^2}{|x|^2}dV_{g_{k}},\) 
where \(g_{k,ij}(x)=g_{0,ij}(r_kx)\) .
Taking the limit, we obtain
\(\int _{B_r\setminus B_{rL^{-1}}}\frac{{u^{\prime }}^2}{|x|^2}dx\le \frac{1}{L}\int _{B_{rL^{-1}}\setminus B_{rL^{-2}}}\frac{{u^{\prime }}^2}{|x|^2}dx.\) 
Letting \(r\)  be sufficiently large in the above inequality, we get \(a=0\) .
Since \(\int _{\partial B_1}\log u^{\prime }=0\) , \(b=1\) .
If we change coordinates: \(x\rightarrow \frac{x}{|x|^2}\) , then \({u^{\prime }}^\frac{4}{n-2}g_{euc}\) 
is just \(g_{euc}\)  in the new coordinates. The convergence of \(d_{g_k}\)  follows
from Proposition REF  directly.
\(\hfill \Box \)
In the above lemma, we fail to write \(c_k\)  in terms of \(r_k\) . Hence the behaviors of
\(u(r_kx)\)  are not really understood. However, the lemma is already strong enough
to help us get the following decays:
Corollary 3.2 
For any \(\tau \in (0,1)\) , there exists \(\delta \) , such that
for any \(r<\delta \) , the following hold
\((1-\tau )\frac{1}{2^n}\le \frac{\mathrm {vol}(B_{2r}\setminus B_r,g)}{\mathrm {vol}(B_r\setminus B_{r/2},g)}\le \frac{1}{2^n}(1+\tau );\\(1-\tau )(2^n-1)\le \frac{\mathrm {vol}(B_{2r}\setminus B_r)}{V_n (d_g(2rx_0,rx_0))^n}\le (2^n-1)(1+\tau ),\,\,\,\,\forall x_0\in \partial B;\\\frac{\int _{B_{r}\setminus B_{r/2}}|x|^\beta |\nabla _{g_0} u|^\alpha dV_{g_0}}{\int _{B_{2r}\setminus B_{r}} |x|^\beta |\nabla _{g_0} u|^\alpha dV_{g_0}}<2^{(n-1)\alpha -n-\beta }(1+\tau ),\,\,\,\,\forall \alpha \in [1,n), \,\,\,\,\beta \in \mathbb {R};\\\frac{\int _{B_{r}\setminus B_{r/2}}|x|^\beta u^\alpha dV_{g_0}}{\int _{B_{2r}\setminus B_{r}} |x|^\beta u^\alpha dV_{g_0}}<2^{(n-2)\alpha -n-\beta }(1+\tau ),\,\,\,\,\forall \alpha \in [1,+\infty ),\,\,\,\,\beta \in \mathbb {R};\\\frac{1}{2^{n-2}}(1-\tau )\le \frac{\int _{\partial B_r}|\nabla _{g_0}\log u|dS_{g_0}}{\int _{\partial B_{2r}}|\nabla _{g_0}\log u|dS_{g_0}}\le \frac{1}{2^{n-2}}(1+\tau );\\2^{-1}(1-\tau )\le \frac{d_g(2rx_0,rx_0)}{d_g(rx_0,rx_0/2)}\le 2^{-1}(1+\tau ),\,\,\,\,\forall x_0\in \partial B;\\(1-\tau )\le \frac{d_g(rx_0,\partial B_{2r})}{d_g(rx_0,2rx_0)}\le (1+\tau ),\,\,\,\,\forall x_0\in \partial B;\\2^{-1}(1-\tau )\le \frac{diam(\partial B_{2r})}{diam(\partial B_r)}\le 2^{-1}(1+\tau ).\) 

Let \(u_k=c_kr_k^\frac{n-2}{2}u_k(r_kx)\)  be as in the proof of Lemma REF , which converges to \(u^{\prime }=|x|^{2-n}\) 
weakly in \(W^{2,p}_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\)  for any \(p<\frac{n}{2}\) .
Then we may assume \(\nabla u_k\)  converges in \(L^q_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\) 
for any \(q<n\) , and \(u_k\)  converges in \(L^q_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\)  for any
\(q\) . By the trace inequality, we can also assume
\(\log u_k\)  converges in \(L^1(\partial B_t)\) .
Now, we prove (REF ):
assume it is not true, then there exists \(r_k\rightarrow 0\) , such that
\(\frac{\mathrm {vol}(B_{2r_k}\setminus B_{r_k},g)}{\mathrm {vol}(B_{r_k}\setminus B_{r_k/2},g)}> \frac{1}{2^n}(1+\tau ),\) 
which means that
\(\frac{\mathrm {vol}(B_{2}\setminus B_1,g_k)}{\mathrm {vol}(B_{1}\setminus B_{1/2},g_k)}> \frac{1}{2^n}(1+\tau ),\) 
where \(g_{k,ij}=u_k^\frac{4}{n-2}g_{0,ij}(r_kx)\) . Letting \(k\rightarrow +\infty \) ,
we get
\(\frac{\mathrm {vol}(B_{2}\setminus B_1,g_\infty )}{\mathrm {vol}(B_{1}\setminus B_{1/2},g_\infty )}\ge \frac{1}{2^n}(1+\tau ),\) 
where \(g_\infty =|x|^{2-n}g_{euc}\) . A contradiction.
Since the proofs to other inequalities are almost the same, we omit them.
\(\hfill \Box \)
The inequalities ()-() will be used to estimate quantities on \(B_{2r}\setminus B_r\) . For example, by (REF ),
\(\mathrm {vol}(B_{2^kr}\setminus B_{2^{k-1}r},g)\le \left(\frac{1+\tau }{2^n}\right)^k\mathrm {vol}(B_{2r}\setminus B_r,g),\) 
then
\(\mathrm {vol}(B\setminus B_r,g)\le C\mathrm {vol}(B_{2r}\setminus B_{r},g).\) 
We give several more applications.
Corollary 3.3 
\((\overline{B}\setminus \lbrace 0\rbrace ,g)\)  is complete.

Let \(\lbrace x_k\rbrace \)  be a sequence which does not contain a convergent subsequence. Then
\(x_k\rightarrow 0\) . Fix an \(x_0\in \partial B\) . For our purposes, it suffices to show
\(d_g(x_k,x_0)\rightarrow +\infty \) .
By () and (), \(d_g(x_k,2x_k)\rightarrow +\infty \) ,
and
\(d_g(x_0,x_k)\ge d_g(x_k,\partial B_{2|x_k|})\ge (1-\tau )d_g(x_k,2x_k)\rightarrow +\infty .\) 
\(\hfill \Box \)
Corollary 3.4 
We have
\(\lim _{r\rightarrow +\infty }\frac{\mathrm {vol}(B_r^g(x_0),g)}{V_nr^n}=1\) 
and \((\overline{B}\setminus \lbrace 0\rbrace ,r^{-2}g,x_0)\) 
converges to \(\mathbb {R}^n\)  in the Gromov-Hausdorff distance
for any \(x_0\) .

Since the proofs for volume convergence and Gromov-Hausdorff
convergence are almost the same, we only prove the volume convergence
here.
Let \(\rho _k\rightarrow +\infty \) , and set \(x_k=r_kx_0\) , such that
\(d_g(x_0,x_k)=\rho _k.\) 
Put \(y_k=\sigma x_k\) , where \(\sigma =2^m\)  is sufficiently large.
We denote
\(\tau _k=d_g(x_k,y_k).\) 
By (),
\(d_g(x_0,y_k)<3d_g(y_k,\frac{1}{2}y_k):=\sigma _k,\) 
hence
\(\tau _k-\sigma _k\le \rho _k\le \tau _k+\sigma _k.\) 
Since
\(d_g(x,x_0)\le d_g(x,y_k)+d_g(x_0,y_k),\) 
for any \(x\in B_{\tau _k-2\sigma _k}^g(y_k)\) ,
\(d_g(x,x_0)\le \tau _k-2\sigma _k+\sigma _k\le \rho _k.\) 
Similarly, for any \(x\in B_{\rho _k}^g(x_0)\)
\(d_g(x,y_k)\le d_g(x,x_0)+d_g(x_0,y_k)\le \tau _k+2\sigma _k.\) 
Then
\(B^g_{\tau _k-2\sigma _k}(y_k)\subset B^g_{\rho _k}(x_0),\,\,\,\,B^g_{\rho _k}(x_0)\subset B^g_{\tau _k+2\sigma _k}(y_k).\) 
Let \(u_k=u(r_kx)r^\frac{n-2}{2}c_k\) , where \(c_k\)  is as in Lemma REF . Since
\(u_k\)  converges to \(|x|^{2-n}\) , and
\(\frac{\sigma _k}{\tau _k}\rightarrow \frac{3}{\sigma -1},\) 
by choosing \(\sigma \)  sufficiently large, for a fixed \(\epsilon \)  and
sufficiently large \(k\) ,
\(B^g_{(1-\epsilon )\tau _k}(y_k)\subset B^g_{\rho _k}(x_0)\subset B^g_{(1+\epsilon )\tau _k}(y_k).\) 
Next, we estimate
\(\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k),g)}{(\lambda \tau _k)^n}.\) 
By (), () and (), we have
\(\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{{(\lambda \tau _k)}^n}&\le &\frac{\mathrm {vol}(B_{{\lambda \tau _k}}^g(y_k),g)}{{(\lambda \tau _k)}^n}\\&=&\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)+\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\setminus B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)+\mathrm {vol}(B\setminus B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)+C\mathrm {vol}(B_{2\sigma r_k}\setminus B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{(\lambda \tau _k)^n}+C\frac{d_g^n(2\sigma r_kx,\sigma r_kx)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{(\lambda \tau _k)^n}+C\frac{\sigma _k^n}{\lambda ^n\tau _k^n}.\) 
It is easy to check that
\(\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\rightarrow \frac{\mathrm {vol}(B^{g_\infty }_{\lambda (1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma },g_\infty )}{(\lambda (1-\sigma ^{-1}))^n}.\) 
To calculate \(\mathrm {vol}(B^{g_\infty }_{(1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma },g_\infty )\) , we use the coordinates change
\(B\rightarrow \mathbb {R}^n\setminus B: \,\,\,\,x\rightarrow z=\frac{x}{|x|^2}.\) 
In the new coordinates, \(g_\infty =g_{euc}\) , and
\(B^{g_\infty }_{\lambda (1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma }=B_{\lambda (1-\sigma ^{-1})}(\sigma ^{-1}\frac{x_0}{|x_0|})\setminus B_{\sigma ^{-1}}.\) 
Thus
\(\mathrm {vol}(B^{g_\infty }_{\lambda (1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma },g_\infty )=V_n((\lambda (1-\sigma ^{-1}))^n-\sigma ^{-n}).\) 
Then we can select \(\sigma \)  to be sufficiently large, such that
\(1-\epsilon \le \frac{\mathrm {vol}(B_{(1+\epsilon )\tau _k}^g(y_k),g)}{V_n((1+\epsilon )\tau _k)^n},\,\,\,\,\frac{\mathrm {vol}(B_{(1-\epsilon )\tau _k}^g(y_k),g)}{V_n((1-\epsilon )\tau _k)^n}\le (1+\epsilon )\) 
when \(k\)  is sufficiently large. By (REF ) and (REF ),
\(1-C\epsilon \le \frac{\mathrm {vol}(B_{\rho _k}(x_0))}{V_n\rho _k^n}\le 1+C\epsilon \) 
when \(k\)  is sufficiently large.
\(\hfill \Box \)
Corollary 3.5 
Assume \((M,g)\)  is conformal compact with finite \(\Vert R\Vert _{L^\frac{n}{2}}\) .
If \(Ric\ge 0\) , and if \((M,g)\)  is complete and noncompact, then \((M,g)=\mathbb {R}^n\) .

By Bishop-Gromov Theorem, \(\mathrm {vol}(B_r^g(x),g)\le V_nr^n\) . Then by a result in [6]}, there exists a compact manifold \((M_0,g_0)\)  and a finite set \(A\subset M_0\) , such that \((M,g)\)  is conformal to \((M_0\setminus A,g_0)\) .
By Corollary REF , \(A\) 
contains a single point, so the corollary follows from the Bishop-Gromov Theorem.
\(\hfill \Box \)
Next, we derive a stronger version of Lemma REF  and
finish the proof of Theorem REF :
Proposition 3.6 \(w=G^{-1}u\)  is in \(W^{2,p}(B)\)  for any \(p\in [1,\frac{n}{2})\) , where \(G\)  is the Green function defined by
\(-\Delta _{g_0}G=\delta _0,\,\,\,\,G|_{\partial B}=0.\) 

By a direct computation,
\(\Delta _{g_0} u=G\Delta _{g_0} w+2\nabla _{g_0} G\nabla _{g_0} w=-c(n)Ru^\frac{n+2}{n-2}+c(n)R(g_0)u.\) 
Then
\(-\Delta _{g_0} w&=&c(n)G^{-1}Ru^\frac{n+2}{n-2}+2\nabla _{g_0}\log G\nabla _{g_0} w-c(n)R(g_0)w\\&=&c(n)Ru^\frac{4}{n-2}w+2\nabla _{g_0}\log G\nabla _{g_0} w-c(n)R(g_0)w\\&:=&f.\) 
It is well known that \(G=r^{2-n}(1+O(1))\)  near 0. By the gradient estimate
of Cheng and Yau on \(B_{|x|/2}^{g_0}(x)\) ,
\(|\nabla _{g_0}\log G|(x)\le \frac{C}{|x|}\) 
when \(x\)  is small.
First, we show \(w\in L^q\)  for any \(q\) . Indeed, applying () to \(\alpha =q\) , \(\beta =(n-2)q\) ,
we get
\(\int _{B_t}|w|^q\le C\sum _i\int _{B_{2^{-i}t}}|x|^{p(n-2)}|u|^q<C\sum _i((1+\tau )2^{-n})^i<+\infty .\) 
Next, we show \(w\in W^{1,p}(B)\)  for any \(p<n\) . Since
\(|\nabla _{g_0}w|\le G^{-1}|\nabla _{g_0}u|+uG^{-1}|\nabla _{g_0}\log G|\le C(|x|^{n-2}|\nabla _{g_0}u|+|x|^{n-3}u),\) 
we may apply () to \(\alpha =p\)  and \(\beta =(n-3)p\)  to
get
\(\int _{B_t}(|x|^{n-3}|u|)^p<+\infty \) 
and () to \(\alpha =p\)  and \(\beta =(n-2)p\)  to obtain
\(\int _{B_t}|x|^{n-2}|\nabla u|^p<+\infty .\) 
Then \(\int _{B_t}|\nabla w|^p<+\infty \)  for any \(p<n\) . Let \(\varphi \in \mathcal {D}(B)\) , and \(\eta _\epsilon \)  be a cutoff function which is
1 in \(B\setminus B_{2\epsilon }\) , 0 in \(B_\epsilon \) , and satisfies \(|\nabla \eta _\epsilon |<\frac{C}{\epsilon }\) . It is essay to check that \(\eta _\epsilon w\)  is bounded in \(W^{1,p}(B)\) , hence a subsequence of \(\eta _\epsilon w\) 
converges weakly in \(W^{1,p}(B)\) . Obviously, \(w\)  is the limit, hence \(w\in W^{1,p}(B)\) .
Next, we show \(f\in L^p\)  for any \(p<\frac{n}{2}\)  and \(w\)  solves the equation \(-\Delta _{g_0}w=f\)  weakly in \(B\) .
Since
\(|f|\le C(|R|u^\frac{4}{n-2}w+|x|^{-1}|\nabla _{g_0}w|+w),\) 
by the fact that \(|R|u^\frac{4}{n-2}\in L^\frac{n}{2}\) , \(w\in L^q\)  for any \(q>0\)  and \(\nabla w\in L^p\)  for any \(p<n\) ,
one is easy to check that \(f\in L^p\)  for any \(p<\frac{n}{2}\) . Then
\(\int _{B}\nabla _{g_0}\varphi \nabla _{g_0} wdV_{g_0}=\lim _{\epsilon \rightarrow 0}\int _B\nabla _{g_0}\eta _\epsilon \varphi \nabla _{g_0} wdV_{g_0}=\lim _{\epsilon \rightarrow 0}\int _B\eta _\epsilon \varphi f=\int _B\varphi f,\) 
hence \(w\)  is a weak solution.
By the theory of elliptic equations, we can complete the proof
without difficulty.
\(\hfill \Box \)

conformally immersed submanifolds in \(\mathbb {R}^{n+k}\)
In this section, we let \(F:(B\setminus \lbrace 0\rbrace ,g_0) \rightarrow \mathbb {R}^n\)  be a conformal immersion with
\(\Vert A\Vert _{L^n}<+\infty \) , where \(A\)  is the second fundamental form. We set
\(g=F^*(g_0)=u^\frac{4}{n-2}g_0.\) 
Obviously
\(\int _{B}|R|^\frac{n}{2}<+\infty .\) 
By result in [3]}, \((B\setminus \lbrace 0\rbrace ,g)\)  can be extended
to a compact metric space in the case when \(\mathrm {vol}(B\setminus \lbrace 0\rbrace ,g)<+\infty \) . We assume \(\mathrm {vol}(F(B\setminus \lbrace 0\rbrace ))=+\infty \)  in this section. It follows from
Corollary REF  that \((\overline{B}_\frac{1}{2}\setminus \lbrace 0\rbrace ,g)\)  is complete.
The goal of this section is to prove Theorem REF .
Proof of Theorem REF : First of all,
we can find \(c_k\) , such that \(c_k r_k^\frac{n-2}{2} u(r_k x)\) 
converges to \(|x|^{2-n}\)  weakly in \(W^{2,p}_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\) 
for any \(p<\frac{n}{2}\) .
Set
\(F_k=c_k^\frac{2}{n-2}(F(r_kx)-F(r_kx_0)).\) 
Since
\(\left|\frac{\partial F_k}{\partial x^i}\right|^2=r_k^2c_k^\frac{4}{n-2}(u_k(r_kx))^\frac{4}{n-2}=(c_kr_k^\frac{n-2}{2}u_k(r_kx))^\frac{4}{n-2},\) 
\(|\nabla F_k|\)  is bounded in \(W^{2,p}(B_r\setminus B_\frac{1}{r})\)  by Lemma REF . Thus, we may assume
\(F_k\)  converges weakly in \(W^{3,p}_{loc}\)  to a map \(f_\infty \)  with
\(F_\infty (\mathbb {R}^n)\subset \mathbb {R}^n\)  and
\(\frac{\partial F_\infty }{\partial x^i}\frac{\partial F_\infty }{\partial x^j}=|x|^{-4}\delta _{ij}.\) 
It will be convenient if we change the coordinates
\(x\rightarrow y=\frac{x}{|x|^2}.\) 
In new coordinates,
\(g_\infty (y)=g_{euc}(y).\) 
Then \(F_\infty \)  can be considered as a isometric map from \(\mathbb {R}^n\setminus \lbrace 0\rbrace \) 
to itself.
Let \(\gamma (t)=ty\) , where \(y\in \mathbb {S}^{n-1}\) . Since \(\gamma (t)\)  is a geodesic
in \(\mathbb {R}^n\setminus \lbrace 0\rbrace \) , \(f(\gamma (t))\)  should be a ray of \(\mathbb {R}^n\) .
In addition, it is easy to check that
\(d_{g_\infty }(F_\infty (y^{\prime }),F_\infty (y^{\prime \prime }))\rightarrow 0\)  as \(y^{\prime }\) , \(y^{\prime \prime }\) 
approach 0, then \(\lim \limits _{y\rightarrow 0}f_\infty (y)\)  exists.
Without loss of generality, we assume the limit is 0. Then
\(F_\infty (\gamma (t))=tF_\infty (y)\) .
Since \(F_\infty \)  is conformal, for any \(X\in T\partial B_1=\mathbb {S}^{n-1}\)  with \(|X|=1\) , we have
\(F_{\infty ,*}(X)\bot F_{\infty ,*}(\frac{\partial }{\partial r}),\,\,\,\,|F_{\infty ,*}(X)|=1.\) 
Thus, \(F_\infty |_{ \mathbb {S}^{n-1}}\)  is an isometric map from
\(\mathbb {S}^{n-1}\)  to \(\mathbb {S}^{n-1}\) . Since \(\mathbb {S}^{n-1}\)  is simple-connected, \(F_\infty |_{\mathbb {S}^{n-1}}\)  is a homeomorphism, as follows from the fact that
an isometric map is a covering map. Therefore, we may assume \(F_\infty (y)=y\) .
Consequently, in the origin coordinates, we have
\(F_\infty (x)=\frac{x}{|x|^2}.\) 
Now, we let \(x_k\rightarrow 0\) .
Assume \(a=|x_0|\) , and \(x_k^{\prime }=a\frac{x_k}{|x_k|}\) . Put \(r_k=|x_k|\) , \(y_k=\sigma x_k\) , where \(\sigma =2^{-m}\)  is sufficiently large. Since
\(d_g(\sigma x_k,x_k^{\prime })\le Cd_g(\sigma x_k,2\sigma x_k),\,\,\,\,and \,\,\,\,\frac{d_g(\sigma x_k,2\sigma x_k)}{d_g(x_k,\sigma x_k)}\rightarrow \frac{1}{\sigma },\) 
when \(\sigma \)  is sufficiently large,
\(1-\epsilon \le \frac{d_g(x_k,x_0)}{d_g(x_k,\sigma x_k)}\le 1+\epsilon .\) 
By an arguments as in the proof of Lemma REF , for any \(\tau \in (0,1)\) ,
\(\frac{1-\tau }{2}\le \frac{|F(rx)-F(rx/2)|}{|F(rx/2)-F(rx/4)|}\le \frac{1+\tau }{2}\) 
for any \(x\in \partial B\)  and sufficiently small \(r\) , then for sufficiently large \(k\) ,
\(1-\epsilon \le \frac{|F(x_k)-F(x_0^{\prime })|}{|F(x_k)-F(\sigma x_k)|}<1+\epsilon .\) 
Since
\(\frac{|F(x_k)-F(\sigma x_k)|}{d_g(x_k,\sigma x_k)}=\frac{|F_k(x_k/|x_k|)-F_k(\sigma x_k/|x_k|)|}{d_{g_k}(x_k/|x_k|,\sigma x_k/|x_k|)}\rightarrow 1,\) 
\(1-C\epsilon <\frac{|F(x_k)-F(x_0)|}{d_g(x_k,x_0)}\le 1+C\epsilon \) 
when \(k\)  is sufficiently large.
\(\hfill \Box \)

 4 dimensional Gauss-Bonnet-Chern formulas
In this section, we assume \(n=4\)  and discuss the Gauss-Bonnet-Chern formulas.
For our purpose, we have to estimate \(\int _{B_{t_1}\setminus B_{t_2}}\Delta _gRdV_g\)  and \(\int _{B_{t_1}\setminus B_{t_2}}\Delta _{g_0}^2\log udV_{g_0}\) , or \(\int _{\partial B_t}n_{\partial B_t,g}(R)dS_{g}\)  and
\(\int _{\partial B_t}\frac{\partial \Delta _{g_0}\log u}{\partial r}dS_{g_0}\) .
For simplicity, we put \(dS_{g_0}=\Theta (r,\theta )d\mathbb {S}^3\) , \(\phi =\log u\)  and
\(F_1(r)=\int _{\mathbb {S}^3}R(r,\theta )u^{2}(r,\theta )\Theta (r,\theta )d\mathbb {S}^3,\,\,\,\,H_1=-\int _{S^3}R(r,\theta )\frac{\partial }{\partial r}(u^{2}(r,\theta )\Theta (r,\theta ))d\mathbb {S}^3\) 
\(F_2(r)=\int _{S^3}\Delta _{g_0}\phi (r,\theta )\Theta (r,\theta )d\mathbb {S}^3,\,\,\,\,H_2=-\int _{S^3}\Delta _{g_0}\phi \frac{\partial }{\partial r}\Theta (r,\theta )d\mathbb {S}^3.\) 
Note that
\(\int _{\partial B_r}n_{g,\partial B_r}(R)dS_g=\int _{S^3}u^2\frac{\partial R}{\partial r}\Theta d\mathbb {S}^3=F_1^{\prime }(r)+H_1(r),\) 
and
\(\int _{\partial B_r}\frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}=F_2^{\prime }(r)+H_2(r).\) 
The key point is that we will find \(r_k\rightarrow 0\) , such that \(F_i^{\prime }(r_k)+H_i(r_k)\)  converges.
The following lemma will play a very important role in the following discussion.
Lemma 5.1 
Let \(f\in C^1[\frac{r_0}{4}, 2r_0]\) , \(h\in C^0[\frac{r_0}{4}, 2r_0]\)  and \(b_1\) , \(b_2\)  are constants. Assume
\(\left|\int _{\frac{r_0}{4}}^{\frac{r_0}{2}}fdt-\frac{3}{32}b_1r_0^2\right|+\left|\int _{r_0}^{2r_0}fdt-\frac{3}{2}b_1r_0^2\right|\le ar_0^2,\,\,\,\,\int _{\frac{r_0}{4}}^{2r_0}|h-b_2|<ar_0.\) 
Then there exists \(\xi \in [r_0/4,2r_0]\)  such that
\(|f^{\prime }(\xi )+h(\xi )-b_1-b_2|\le 12a.\) 

Since we can replace \(f\)  with \(f-b_1r\)  and \(h\)  with \(h-b_2\) , it suffices for our aim to prove the case when \(b_1=b_2=0\) .
By The Mean Value Theorem for Integrals, there exists \(\xi _1\in [r_0/4,r_0/2]\)  and \(\xi _2\in [r_0,2r_0]\) , such that
\(\frac{r_0}{4}f(\xi _1)=\int _\frac{r_0}{4}^\frac{r_0}{2}f(t)dt,\,\,\,\,r_0f(\xi _2)=\int _{r_0}^{2r_0}f(t)dt,\) 
which yields that
\(|f(\xi _1)|\le 4ar_0,\,\,\,\,|f(\xi _2)|\le ar_0.\) 
Then
\(\left|\int _{\xi _1}^{\xi _2}(f^{\prime }+h)\right|\le |f(\xi _1)-f(\xi _2)|+\int _{\xi _1}^{\xi _2}|h|\le 5ar_0+\int _{r_0/4}^{2r_0}|h|\le 6ar_0.\) 
Using the Mean Value Theorem for Integrals again, we can find \(\xi \in [\xi _1,\xi _2]\) , such that
\((\xi _2-\xi _1)|f^{\prime }(\xi )+h(\xi )|\le 6a r_0.\) 
Noting that \(r_0/2<\xi _2-\xi _1\) , we complete the proof.
\(\hfill \Box \)
Lemma 5.2 For any sufficiently small \(r\) , we have
\(\left|\int _r^{2r}F_1(t)dt\right|<\alpha (r)r^2, \,\,\,\,\int _r^{2r}|H_1|(t)dt<\alpha (r)r,\) 
and
\(\left|\int _r^{2r}F_2(t)dt+6\omega _3r^2\right|<\alpha (r)r^2, \,\,\,\,\int _r^{2r}|H_2-12\omega _3|(t)dt<\alpha (r)r,\) 
where \(\lim \limits _{r\rightarrow 0}\alpha (r)=0\) , and \(\omega _3=2\pi ^2\)  is the volume of the 3-dimensional sphere.

We have
\(\left|\int _{r}^{2r} F_1(t)\right|& \le & \int _{r}^{2r} \int _{S^3} |R(g)| u^2 \Theta dSdt \\& =& \int _{B_{2r}\setminus B_r}|R(g)| u^2dV_{g_0}\\&\le &\left(\int _{B_{2r} \setminus B_{r}} | R(g) u^2 |^2 dV_{g_0}\right)^{\frac{1}{2}}\left(\int _{B_{2r} \setminus B_{r}} dV_{g_0}\right)^{\frac{1}{2}} \\&\le & C\Vert R\Vert _{L^2(B_{2r},g)}r^2.\) 
and
\(\int _{r}^{2r} |H_1(t)| dt& \le & 2\int _{r}^{2r} \int _{\mathbb {S}^3}|R(g)| u^2 \frac{\partial \log u}{\partial t}\Theta d\mathbb {S}^3dt+ 2\int _{r}^{2r} \int _{\mathbb {S}^3}|R(g)| u^2\frac{\partial \log \Theta (t,\theta )}{\partial t}(\Theta d\mathbb {S}^3dt) \\&\le & C(\Vert R\Vert _{L^2(B_{2r},g)}\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}+\int _{B_{2r}\setminus B_r}|R(g)| u^2\frac{1}{r}dV_{g_0})\\&\le & C(\Vert R\Vert _{L^2(B_{2r},g)}\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}+Cr\Vert R(g)\Vert _{L^2(B_{2r},g)}.\) 
By Lemma REF ,
\(\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}\le Cr\Vert Ru^2+R(g_0)\Vert _{L^2(B_{2r}\setminus B_r,g_0)}\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r+r^2),\) 
hence
\(\int _r^{2r}|H_1(t)|dt\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r+r^2).\) 
Next, we discuss \(F_2\) . Since
\(-\Delta _{g_0}\phi -|\nabla _{g_0}\phi |^2=c(n)Ru^2-c(n)R(g_0),\) 
we obtain
\(\int _{r}^{2r} F_2(t) dt& =& \int _{B_{2r}\setminus B_r}\Delta _{g_0}\phi dV_{g_0} \\& =& -\int _{B_{2r}\setminus B_r}|\nabla _{g_0} \phi |^2dV_{g_0}-\int _{B_{2r}\setminus B_r}(R(g) u^2-R(g_0))dV_{g_0}.\) 
Note that
\(\int _{B_{2r}\setminus B_r}|R(g) u^2-R(g_0)|dV_{g_0}\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r^2+r^4).\) 
To get the estimate of \(\int _{r}^{2r}F_2\) , we only need to show that
\(\lim _{r\rightarrow 0}\frac{1}{r^2}\int _{B_{2r}\setminus B_r}|\nabla _{g_0}\phi |^2dV_{g_0}=6\omega _3.\) 
Assume there exists \(r_k\rightarrow 0\) , such that
\(\lim _{k\rightarrow \infty }\frac{1}{r_k^2}\int _{B_{2r_k}\setminus B_{r_k}}|\nabla _{g_0}\phi |^2dV_{g_0}=\lambda \ne 6\omega _3.\) 
Set \(u_k=c_kr_ku(r_kx)\) , where \(c_k\)  is chosen such that \(\int _{\partial B_1}\log u_k=0\) .
By the arguments in Section 3, \(\log u_k(x)\)  converges to
\(-\log |x|^{-2}\)  weakly in \(W^{2,p}_{loc}(\mathbb {R}^4\setminus \lbrace 0\rbrace )\) . Then, after passing to a subsequence,
\(\int _{B_2\setminus B_1}|\nabla _{g_0(r_kx)}\log u_k|^2\)  converges to \(6\omega _3\) , hence
\(\lim _{k\rightarrow +\infty }\frac{1}{r_k^2}\int _{B_{2r_k}\setminus B_{r_k}}|\nabla _{g_0}\phi |^2dV_{g_0}=\lim _{k\rightarrow +\infty }\int _{B_2\setminus B_1}|\nabla _{g_0(r_kx)}\log u_k|^2dV_{g_0(r_kx)}=6\omega _3.\) 
A contradiction.
Lastly, we calculate \(\int _r^{2r}|H_2-12\omega _3|\) :
\(\int _{r}^{2r} |H_2(t) -12\omega _3|dt\le \int _{B_{2r}\setminus B_r}\left||\nabla _{g_0} \phi |^2\frac{\partial \log \Theta }{\partial r}-\frac{12}{\Theta }\right|dV_{g_0}+C\int _{B_{2r}\setminus B_r}|R(g) u^2-R(g_0)|\frac{1}{r}dV_{g_0}.\) 
The same argument as above shows that
\(\lim _{r\rightarrow 0}\frac{1}{r}\int _{B_{2r}\setminus B_r}\left||\nabla _{g_0} \phi |^2\frac{\partial \log \Theta }{\partial r}-\frac{12}{\Theta }\right|dV_{g_0}=0.\) 
This completes the proof.
\(\hfill \Box \)
We will give several applications here. First, we study the integrability of \(\Delta _{g} R\) :
Lemma 5.3 There exists \(r_k \rightarrow 0\)  such that
\(\int _{B_{\frac{1}{2}}\setminus B_{r_k}}\Delta _g R(g) dV_g\rightarrow \int _{\partial B_\frac{1}{2}}\frac{\partial R}{\partial r}dS_g.\) 
Here \(B_r\)  is a ball with respect to \(g_0\) .

We have
\(\int _{B_{\frac{1}{2}}\setminus B_{r}}\Delta _g R(g) dV_g=\int _{\partial B_\frac{1}{2}}n_{g,\partial B_\frac{1}{2}}(R)dS_g-\int _{\partial B_r}n_{g,\partial B_r}(R)dS_g.\) 
Applying Lemma REF  to \(b_1=b_2=0\)  and \(f=F_1\) , \(h=H_1\)  , we deduce this lemma from (REF ).
\(\hfill \Box \)
We recall some basic properties of \(Q\) -curvatures, c.f. [8]}, [9]}. On a 4 dimensional manifold, the Paneitz operator is defined as follows:
\(P_{g_0}\varphi =\Delta _{g_0}^2\varphi +div_{g_0}\left(\frac{2}{3}R_{g_0}\nabla _{g_0}\varphi -2Ric_{g_0}^{ij}\varphi _i\frac{\partial }{\partial x^j}\right).\) 
The \(Q\) -curvature of \(g\)  satisfies the following equations:
\(Q(g)=-\frac{1}{12}\Delta _gR(g)-\frac{1}{4}|Ric(g)|^2+\frac{1}{12}R^2,\) 
\(P_{g_0}\phi +2Q(g_0)=2Q(g)e^{4\phi }.\) 
For simplicity, we define
\(T(\phi )=\frac{1}{3}R_{g_0}\frac{\partial \phi }{\partial r}-Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r}).\) 
Lemma 5.4 
There exists \(r_k\rightarrow 0\) , such that
\(\lim _{k\rightarrow +\infty }\int _{B_\frac{1}{2}\setminus B_{r_k}} Q(g)dV_g=\int _{B_\frac{1}{2}}Q(g_0)dV_{g_0}-4\omega _3+\int _{\partial B_\frac{1}{2}}\left(\frac{1}{2}\frac{\partial \Delta _{g_0}\phi }{\partial r}+T(\phi )\right).\) 

We have
\(\int _{B_\frac{1}{2}\setminus B_r}Q_gdV_g&=&\int _{B_\frac{1}{2}\setminus B_r}Q_{g_0}dV_{g_0}+\frac{1}{2}\int _{B_\frac{1}{2}\setminus B_r}P_{g_0}(\phi )dV_{g_0}\\&=&\int _{B_\frac{1}{2}\setminus B_r}Q_{g_0}dV_{g_0}\\&&+\frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\left(\frac{\partial \Delta _{g_0}\phi }{\partial r}+\frac{2}{3}R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}.\) 
First, we check that
\(\int _{\partial B_r}\left(\frac{2}{3}R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}\rightarrow 0.\) 
By (),
\(\lim _{r\rightarrow 0}\int _{\partial B_r}|\nabla _{g_0}\phi |dS_{g_0}= 0.\) 
Then (REF ) follows.
Then, by applying Lemma REF  to \(f=F_2/2\) , \(h=H_2/2\)  and \((b_1,b_2)=(-2\omega _3,6\omega _3)\) , we deduce from (REF )
that there exists \(r_k\) , such that
\(\int _{\partial B_{r_k}}\frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\rightarrow 8\omega _3.\) 
Therefore, we complete the proof.
\(\hfill \Box \)
Next, we discuss the relationship between \(\Vert R\Vert _{L^2}\) 
and \(\Vert Riem\Vert _{L^2}\) :
Lemma 5.5 
We have
\(\int _{B_\frac{1}{2}}|Riem(g)|^2dV_g<+\infty .\) 

It is well-known that
\(Riem(g)=W(g)+\frac{1}{2}(Ric(g)-\frac{1}{6}R(g)g){~\wedge \!\!\!\!\!\bigcirc ~}g,\) 
where \(W\)  is the Weyl tensor and \({~\wedge \!\!\!\!\!\bigcirc ~}\)  is the Kulkarni-Nomizu product. Since \(|W|^2dV_g\)  is conformal invariant, we only need to check
\(Ric(g)\in L^2\)  here.
Recall that
\(Q(g)=-\frac{1}{12}\Delta _gR(g)-\frac{1}{4}|Ric(g)|^2+\frac{1}{12}R^2\) ,
which means that
\(\int _{B_\frac{1}{2}\setminus B_r}|Ric(g)|^2dV_g&=&\frac{1}{3}\int _{B_\frac{1}{2}\setminus B_r}R^2dV_{g}-\frac{1}{3}\int _{B_\frac{1}{2}\setminus B_r}\Delta _gR(g)dV_{g}-4\int _{B_\frac{1}{2}\setminus B_r}Q(g)dV_{g}\\&=&\frac{1}{3}\int _{B_\frac{1}{2}}R^2dV_{g}-4\int _{B_\frac{1}{2}}Q(g_0)dV_{g_0}\\&&-\frac{1}{3}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\frac{\partial R}{\partial r}dS_g-2\int _{\partial (B_\frac{1}{2}\setminus B_r)}(\frac{\partial \Delta _{g_0}\phi }{\partial r}+2T(\phi ))dS_{g_0}\) 
Applying Lemma REF  to \(f=\frac{1}{3}F_1+2F_2\) , \(h=\frac{1}{3}H_1+2H_2\) , \((b_1,b_2)=(-8\omega _3,24\omega _3)\) , we can find \(r_k\rightarrow 0\) , such that
\(\frac{1}{3}\int _{\partial B_{r_k}}\frac{\partial R}{\partial r}dS_g+2\int _{\partial B_{r_k}}(\frac{\partial \Delta _{g_0}\phi }{\partial r}+2T(\phi ))\rightarrow 16\omega _3.\) 
 \(\hfill \Box \)
Lastly, we consider the formula for Pfaffian form:
Lemma 5.6 
We have
\(\int _{B_\frac{1}{2}}Pf(g)&=&-4\omega _3+\int _{B_\frac{1}{2}}Pf(g_0)+\frac{1}{2}\int _{\partial B_\frac{1}{2}} \frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\\&&+\frac{1}{12}\int _{\partial B_\frac{1}{2}}u^2\frac{\partial R}{\partial r}dS_{g_0}-\frac{1}{12}\int _{B_\frac{1}{2}}\Delta _{g_0}R(g_0)dV_{g_0}\\&&+\int _{\partial B_\frac{1}{2}}T(\phi )dS_{g_0}.\) 

Recall that (c.f. [9]})
\(Pf(g)=\frac{1}{8}|W(g)|^2+\frac{1}{12}R^2-\frac{1}{4}|Ric(g)|^2,\) 
where \(W\)  is the Weyl tensor. Since
\(\int _{B}|W(g)|^2dV_{g}=\int _B|W(g_0)|^2dV_{g_0}<+\infty ,\) 
\(Pf(g)\)  is integrable. Since
\(\begin{aligned}Pf(g) & = \frac{1}{8} |W(g)|^2 dV_g+ Q(g) dV_g+ \frac{1}{12} \Delta _g R(g) dV_g, \\Pf(g_0) & = \frac{1}{8} |W(g_0)|^2 dV_{g_0}+ Q(g_0) dV_{g_0}+ \frac{1}{12} \Delta _{g_0} R(g_0) dV_{g_0},\end{aligned}\) 
and
\(P_{g_0} \phi + 2Q(g_0) =2 Q(g) e^{4 \phi },\) 
where \(g = u^2 g_0 = e^{2 \phi } g_0\) ,
we have
\(\int _{B_\frac{1}{2}\setminus B_r} Pf(g)& =& \int _{B_\frac{1}{2}\setminus B_r} Pf(g_0)+ \frac{1}{2}\int _{B_\frac{1}{2}\setminus B_r} P_{g_0} \phi dV_{g_0}\\&& + \frac{1}{12} \int _{B_\frac{1}{2}\setminus B_r} \Delta _g R(g) dV_g-\frac{1}{12} \int _{B_\frac{1}{2}\setminus B_r} \Delta _{g_0} R(g_0) dV_{g_0}\\& =&\int _{B_\frac{1}{2}\setminus B_r}Pf(g_0)+ \frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)} \frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\\&&+\frac{1}{12}\int _{\partial (B_\frac{1}{2}\setminus B_r)}u^2\frac{\partial R}{\partial r}dS_{g_0}-\frac{1}{12}\int _{B_\frac{1}{2}\setminus B_r}\Delta _{g_0}R(g_0)dV_{g_0}\\&&+\frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\left(R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}.\) 
Apply Lemma REF  to \(f=\frac{1}{12}F_1+\frac{1}{2}F_2\) ,
and \(h=\frac{1}{12}H_1+\frac{1}{2}H_2\) , \((b_1,b_2)=(-2\omega _3,6\omega _3)\) , which suffices to complete the proof.
\(\hfill \Box \)
Theorem REF , REF  and REF  can be deduced from
Lemma REF , REF  and REF  easily.
